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Find the point where maximum intervals overlap

Last Updated : 23 Jul, 2025
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Consider a big party where a log register for guest's entry and exit times is maintained. Find the time at which there are maximum guests in the party. Given the Entry(Entry[]) and Exit (Exit[]) times of individuals at a place.

Note: Entries in the register are not in sorted order.

Examples: 

Input: Entry[] = [1, 2, 9, 5, 5]
Exit[] = [4, 5, 12, 9, 12]
Output: 3 5
Explanation:
Time Event Type Total Guests Present
------------------------------------------------
1 Entry 1
2 Entry 2
4 Exit 1
5 Entry 2
5 Entry 3 // Max Guests
5 Exit 2
9 Exit 1
10 Entry 2
12 Exit 1
12 Exit 0
The maximum number of guests present at any time is 3, occurring at time 5.

Input: Entry[] = [3, 0, 5, 8, 6]
Exit[] = [7, 6, 9, 10, 8]
Output: 3 6
Explanation: Maximum guests are 3 at time 6

Input: Entry[] = [2, 3, 5, 7, 8]
Exit[] = [4, 6, 8, 9, 10]
Output: 2 5
Explanation: Maximum guests are 2 at time 5

Table of Content

[Naive Approach] - Using Nested for loop - O((max-min+1)*n) Time and O(max-min+1) Space

Find the minimum and maximum times from the guest entry and exit times, then iterate through this range to count the number of guests present at each time. The maximum count gives the peak number of guests.

C++ 
// Program to find maximum guest at any time in a party
#include <bits/stdc++.h>
using namespace std;

vector<int> findMaxGuests(vector<int> &Entry, vector<int> &Exit)
{
    int n = Entry.size();
    int mini_time = INT_MAX, max_time = INT_MIN;
    
    // find the minimum time of the guests
    for (int i = 0; i < n; i++)
    {
        mini_time = min(mini_time, Entry[i]);
    }
    
    // find the maximum time of the guests
    for (int i = 0; i < n; i++)
    {
        max_time = max(max_time, Exit[i]);
    }

    // ith index store the count of the guest 
    // at time (mini_time + i)
    vector<int> count(max_time - mini_time + 1, 0); 
    
    // traverse over the array
    for (int i = 0; i < n; i++)
    {
        for (int j = Entry[i]; j <= Exit[i]; j++)
        {
            // increase the count of the guest at the
            // time they are present
            count[j - mini_time]++;
        }
    }
    // find the maximum guests number at any time and store it
    int time = -1, max_guests = INT_MIN;
    for (int i = mini_time; i <= max_time; i++)
    {
        if (max_guests < count[i - mini_time])
        {
            max_guests = count[i - mini_time];
            time = i;
        }
    }

    return {max_guests, time};
}

// Driver program to test above function
int main()
{
    vector<int> Entry = {1, 2, 10, 5, 5};
    vector<int> Exit = {4, 5, 12, 9, 12};
    int n = Entry.size();
    vector<int> ans = findMaxGuests(Entry, Exit);
    cout << ans[0] << " " << ans[1];
    return 0;
}
Java 
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;

class GfG {

    public static int[] findMaxGuests(int[] Entry,
                                      int[] Exit)
    {
        // Finding maximum Entrying time
        int n = Entry.length;
        int maxa = Arrays.stream(Entry).max().getAsInt();

        // Finding maximum Exiting time
        int maxb = Arrays.stream(Exit).max().getAsInt();

        int maxc = Math.max(maxa, maxb);

        int[] x = new int[maxc + 2];
        Arrays.fill(x, 0);

        int cur = 0, idx = 0;

        // Creating an auxiliary array
        for (int i = 0; i < n; i++) {
            // Lazy addition
            for (int j = Entry[i]; j <= Exit[i]; j++) {
                x[j]++;
            }
        }

        int maxy = Integer.MIN_VALUE;

        // Lazily Calculating value at index i
        for (int i = 0; i <= maxc; i++) {
            cur = x[i];
            if (maxy < cur) {
                maxy = cur;
                idx = i;
            }
        }
        return new int[] { maxy, idx };
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] Entry = new int[] { 1, 2, 10, 5, 5 };
        int[] Exit = new int[] { 4, 5, 12, 9, 12 };
        int[] ans = findMaxGuests(Entry, Exit);
        System.out.println(ans[0] + " " + ans[1]);
    }
}
Python 
def findMaxGuests(Entry, Exit):
    mini_time = min(Entry)
    max_time = max(Exit)
    n = len(Entry)
    count = [0] * (max_time - mini_time + 1)

    for i in range(n):
        for j in range(Entry[i], Exit[i] + 1):
            count[j - mini_time] += 1

    max_guests = max(count)
    time = mini_time + count.index(max_guests)

    return [max_guests, time]


# Driver code
Entry = [1, 2, 10, 5, 5]
Exit = [4, 5, 12, 9, 12]

ans = findMaxGuests(Entry, Exit)
print(ans[0], " ", ans[1])
C# 
using System;

class GfG {
    static int[] findMaxGuests(int[] Entry, int[] Exit)
    {
        int mini_time = int.MaxValue, max_time
                                      = int.MinValue;
        int n = Entry.Length;

        for (int i = 0; i < n; i++) {
            mini_time = Math.Min(mini_time, Entry[i]);
            max_time = Math.Max(max_time, Exit[i]);
        }

        int[] count = new int[max_time - mini_time + 1];

        for (int i = 0; i < n; i++) {
            for (int j = Entry[i]; j <= Exit[i]; j++) {
                count[j - mini_time]++;
            }
        }

        int time = -1, max_guests = int.MinValue;
        for (int i = mini_time; i <= max_time; i++) {
            if (max_guests < count[i - mini_time]) {
                max_guests = count[i - mini_time];
                time = i;
            }
        }

        return new int[] { max_guests, time };
    }

    public static void Main()
    {
        int[] Entry = { 1, 2, 10, 5, 5 };
        int[] Exit = { 4, 5, 12, 9, 12 };

        int[] ans = findMaxGuests(Entry, Exit);
        Console.WriteLine(ans[0] + " " + ans[1]);
    }
}
JavaScript 
function findMaxGuests(Entry, Exit)
{
    let mini_time = Math.min(...Entry);
    let max_time = Math.max(...Exit);
    let n = Entry.length;

    let count = new Array(max_time - mini_time + 1).fill(0);

    for (let i = 0; i < n; i++) {
        for (let j = Entry[i]; j <= Exit[i]; j++) {
            count[j - mini_time]++;
        }
    }

    let time = -1, max_guests = -Infinity;
    for (let i = mini_time; i <= max_time; i++) {
        if (max_guests < count[i - mini_time]) {
            max_guests = count[i - mini_time];
            time = i;
        }
    }

    return [ max_guests, time ];
}

// Driver Code
let Entry = [ 1, 2, 10, 5, 5 ];
let Exit = [ 4, 5, 12, 9, 12 ];

let ans = findMaxGuests(Entry, Exit);
console.log(ans[0], " ", ans[1]);

Output
3 5

Time Complexity: O((max-min+1)*n) where n is the length of the array of guests
Auxiliary Space: O(max-min+1) where min is the minimum time at which any guests enter and max is the maximum time at which any guests exit.

[Better Approach] - Using Sorting - O(n * logn) Time and O(1) Space

Sort the entry and exit times, then use two pointers to track guests arriving and leaving. Maintain a count of guests present at any time and update the maximum count whenever needed.

C++ 
// Program to find maximum guest at any time in a party
#include <bits/stdc++.h>
using namespace std;

vector<int> findMaxGuests(vector<int> &Entry, vector<int> &Exit)
{
    int n = Entry.size();
    
    // Sort arrival and Exit arrays
    sort(Entry.begin(), Entry.end());
    sort(Exit.begin(), Exit.end());

    // guests_in indicates number of guests at a time
    int guests_in = 1, max_guests = 1, time = Entry[0];
    int i = 1, j = 0;

    // Similar to merge in merge sort to process
    // all events in sorted order
    while (i < n && j < n)
    {
        // If next event in sorted order is arrival,
        // increment count of guests
        if (Entry[i] <= Exit[j])
        {
            guests_in++;

            // Update max_guests if needed
            if (guests_in > max_guests)
            {
                max_guests = guests_in;
                time = Entry[i];
            }
            i++; // increment index of arrival array
        }
        else // If event is Exit, decrement count
        {    // of guests.
            guests_in--;
            j++;
        }
    }

    return {max_guests, time};
}

// Driver program to test above function
int main()
{
    vector<int> Entry = {1, 2, 10, 5, 5};
    vector<int> Exit = {4, 5, 12, 9, 12};
    vector<int> ans = findMaxGuests(Entry, Exit);
    cout << ans[0] << " " << ans[1];
    return 0;
}
Java 
import java.util.*;

class GFG {
    static int[] findMaxGuests(int[] Entry, int[] Exit)
    {
        int n = Entry.length;

        // Sort arrival and Exit arrays
        Arrays.sort(Entry);
        Arrays.sort(Exit);

        int guests_in = 1, max_guests = 1, time = Entry[0];
        int i = 1, j = 0;

        // Merge-like processing for sorted events
        while (i < n && j < n) {
            if (Entry[i] <= Exit[j]) {
                guests_in++;

                // Update max_guests if needed
                if (guests_in > max_guests) {
                    max_guests = guests_in;
                    time = Entry[i];
                }
                i++;
            }
            else {
                guests_in--;
                j++;
            }
        }
        return new int[] { max_guests, time };
    }

    // Driver program to test the function
    public static void main(String[] args)
    {
        int[] Entry = { 1, 2, 10, 5, 5 };
        int[] Exit = { 4, 5, 12, 9, 12 };

        int[] ans = findMaxGuests(Entry, Exit);
        System.out.println(ans[0] + " " + ans[1]);
    }
}
Python 
def findMaxGuests(Entry, Exit):

    # Sort arrival and Exit arrays
    Entry.sort()
    Exit.sort()
    n = len(Entry)
    # guests_in indicates number of
    # guests at a time
    guests_in = 1
    max_guests = 1
    time = Entry[0]
    i = 1
    j = 0

    # Similar to merge in merge sort to
    # process all events in sorted order
    while (i < n and j < n):

        # If next event in sorted order is
        # arrival, increment count of guests
        if (Entry[i] <= Exit[j]):

            guests_in = guests_in + 1

        # Update max_guests if needed
            if (guests_in > max_guests):

                max_guests = guests_in
                time = Entry[i]

            # increment index of arrival array
            i = i + 1

        else:
            guests_in = guests_in - 1
            j = j + 1

    return [max_guests, time]


# Driver Code
Entry = [1, 2, 10, 5, 5]
Exit = [4, 5, 12, 9, 12]
n = len(Entry)
ans = findMaxGuests(Entry, Exit)
print(ans[0], " ", ans[1])
C# 
using System;

class GfG {
    static int[] findMaxGuests(int[] Entry, int[] Exit)
    {
        int n = Entry.Length;
        // Sort arrival and Exit arrays
        Array.Sort(Entry);
        Array.Sort(Exit);

        // guests_in indicates number
        // of guests at a time
        int guests_in = 1, max_guests = 1, time = Entry[0];
        int i = 1, j = 0;

        while (i < n && j < n) {

            if (Entry[i] <= Exit[j]) {
                guests_in++;

                // Update max_guests if needed
                if (guests_in > max_guests) {
                    max_guests = guests_in;
                    time = Entry[i];
                }

                // increment index of arrival array
                i++;
            }

            // If event is Exit, decrement
            // count of guests.
            else {
                guests_in--;
                j++;
            }
        }

        return new int[] { max_guests, time };
    }

    // Driver Code
    public static void Main()
    {
        int[] Entry = { 1, 2, 10, 5, 5 };
        int[] Exit = { 4, 5, 12, 9, 12 };
        int n = Entry.Length;
        int[] ans = findMaxGuests(Entry, Exit);
        Console.WriteLine(ans[0] + " " + ans[1]);
    }
}
JavaScript 
function findMaxGuests(Entry, Exit)
{
    // Sort arrival and Exit arrays
    let n = Entry.length;
    Entry.sort((a, b) => a - b);
    Exit.sort((a, b) => a - b);

    let guests_in = 1, max_guests = 1, time = Entry[0];
    let i = 1, j = 0;

    // Process all events in sorted order
    while (i < n && j < n) {
        // If the next event is an arrival, increment guest
        // count
        if (Entry[i] <= Exit[j]) {
            guests_in++;

            // Update max_guests if needed
            if (guests_in > max_guests) {
                max_guests = guests_in;
                time = Entry[i];
            }
            i++; // Move to the next arrival
        }
        else {
            // If it's an Exit, decrement the guest count
            guests_in--;
            j++; // Move to the next Exit
        }
    }

    return [ max_guests, time ];
}

// Sample Input
let Entry = [ 1, 2, 10, 5, 5 ];
let Exit = [ 4, 5, 12, 9, 12 ];

// Function Call
let ans = findMaxGuests(Entry, Exit);
console.log(ans[0], " ", ans[1]);

Output
3 5

[Expected Approach] - Using Difference Array - O(n) Time and O(max-min+1) Space

Use a difference array to record guest entries and exits, then compute a running sum to track the number of guests at any time, updating the maximum count accordingly.

C++ 
#include <bits/stdc++.h>
using namespace std;

vector<int> findMaxGuests(vector<int> &Entry, vector<int> &Exit)
{

    int n = Entry.size();

    // Finding maximum Entrying time O(n)
    int maxa = *max_element(Entry.begin(), Entry.end());

    // Finding maximum Exiting time O(n)
    int maxb = *max_element(Exit.begin(), Exit.end());

    int maxc = max(maxa, maxb);

    int x[maxc + 2];
    memset(x, 0, sizeof x);

    int cur = 0, idx;

    // Creating and auxiliary array O(n)
    for (int i = 0; i < n; i++)
    {
        // Lazy addition
        ++x[Entry[i]];
        --x[Exit[i] + 1];
    }

    int maxy = INT_MIN;

    // Lazily Calculating value at index i O(n)
    for (int i = 0; i <= maxc; i++)
    {
        cur += x[i];
        if (maxy < cur)
        {
            maxy = cur;
            idx = i;
        }
    }
    return {maxy, idx};
}

// Driver code
int main()
{
    vector<int> Entry = {1, 2, 10, 5, 5}, Exit = {4, 5, 12, 9, 12};

    vector<int> ans = findMaxGuests(Entry, Exit);
    cout << ans[0] << " " << ans[1];
    return 0;
}
Java 
import java.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;

class GFG {

    static int[] findMaxGuests(int[] Entry, int[] Exit)
    {
        // Finding maximum Entrying time
        int n = Entry.length;
        int maxa = Arrays.stream(Entry).max().getAsInt();

        // Finding maximum Exiting time
        int maxb = Arrays.stream(Exit).max().getAsInt();

        int maxc = Math.max(maxa, maxb);

        int[] x = new int[maxc + 2];
        Arrays.fill(x, 0);

        int cur = 0, idx = 0;

        // Creating an auxiliary array
        for (int i = 0; i < n; i++) {
            // Lazy addition
            ++x[Entry[i]];
            --x[Exit[i] + 1];
        }

        int maxy = Integer.MIN_VALUE;

        // Lazily Calculating value at index i
        for (int i = 0; i <= maxc; i++) {
            cur += x[i];
            if (maxy < cur) {
                maxy = cur;
                idx = i;
            }
        }
        return new int[] { maxy, idx };
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] Entry = new int[] { 1, 2, 10, 5, 5 };
        int[] Exit = new int[] { 4, 5, 12, 9, 12 };
        int[] ans = findMaxGuests(Entry, Exit);
        System.out.println(ans[0] + " " + ans[1]);
    }
}
Python 
import sys


def findMaxGuests(Entry, Exit):

    n = len(Entry)
    maxa = max(Entry)
    maxb = max(Exit)
    maxc = max(maxa, maxb)
    x = (maxc + 2) * [0]
    cur = 0
    idx = 0

    for i in range(0, n):
        x[Entry[i]] += 1
        x[Exit[i] + 1] -= 1

    maxy = -1
    # Lazily Calculating value at index i
    for i in range(0, maxc + 1):
        cur += x[i]
        if maxy < cur:
            maxy = cur
            idx = i
    return [maxy, idx]


if __name__ == "__main__":

    Entry = [1, 2, 10, 5, 5]
    Exit = [4, 5, 12, 9, 12]

    ans = findMaxGuests(Entry, Exit)
    print(ans[0], " ", ans[1])
C# 
using System;
using System.Linq;

class GfG {

    static int[] findMaxGuests(int[] Entry, int[] Exit)
    {
        int n = Entry.Length;
        // Finding maximum Entrying time
        int maxa = Entry.Max();

        // Finding maximum Exiting time
        int maxb = Exit.Max();

        int maxc = Math.Max(maxa, maxb);

        int[] x = new int[maxc + 2];

        int cur = 0, idx = 0;

        // Creating an auxiliary array
        for (int i = 0; i < n; i++) {
            // Lazy addition
            ++x[Entry[i]];
            --x[Exit[i] + 1];
        }

        int maxy = int.MinValue;

        // Lazily Calculating value at index i
        for (int i = 0; i <= maxc; i++) {
            cur += x[i];
            if (maxy < cur) {
                maxy = cur;
                idx = i;
            }
        }
        return new int[] { maxy, idx };
    }

    // Driver code
    public static void Main()
    {
        int[] Entry = new int[] { 1, 2, 10, 5, 5 };
        int[] Exit = new int[] { 4, 5, 12, 9, 12 };
        int[] ans = findMaxGuests(Entry, Exit);
        Console.WriteLine(ans[0] + " " + ans[1]);
    }
}
JavaScript 
// Javascript implementation of above approach

function findMaxGuests(Entry, Exit)
{
    let n = Entry.length;
    // Finding maximum Entrying time
    let maxa = 0;
    for (let i = 0; i < Entry.length; i++) {
        maxa = Math.max(maxa, Entry[i]);
    }

    // Finding maximum Exiting time
    let maxb = 0;
    for (let i = 0; i < Exit.length; i++) {
        maxb = Math.max(maxb, Exit[i]);
    }

    let maxc = Math.max(maxa, maxb);

    let x = new Array(maxc + 2);
    x.fill(0);

    let cur = 0, idx = 0;

    // Creating an auxiliary array
    for (let i = 0; i < n; i++) {
        // Lazy addition
        ++x[Entry[i]];
        --x[Exit[i] + 1];
    }

    let maxy = Number.MIN_VALUE;

    // Lazily Calculating value at index i
    for (let i = 0; i <= maxc; i++) {
        cur += x[i];
        if (maxy < cur) {
            maxy = cur;
            idx = i;
        }
    }
    return [ maxy, idx ]
}

let Entry = [ 1, 2, 10, 5, 5 ];
let Exit = [ 4, 5, 12, 9, 12 ];
let ans = findMaxGuests(Entry, Exit);
console.log(ans[0], " ", ans[1]);

Output
3 5

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